Is Induction or Deduction the Correct Approach?

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In summary: S(n+1) is true... that S(n+2) is true... that S(n+3) is true... etc.Then you can be sure that S is true for all numbers n, n+1, n+2, n+3 etc.This is because we know that the statement S is true for some number n and then we show that it is also true for the next number. Now that we know it is true for the next number, we can use the same argument to show that it is true for the next number after that, and so on. And because we can keep doing this, it means that the statement S is true for all numbers, not just n but also
  • #1
mtanti
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Here it is, tell me who is wrong...

A definition of Induction and Deduction in a general science field:

Inductive approach- To understand something through obsevation with no prior knowledge and concluding at the end of your observations.

Deductive approach- To understand something through reasoning using prior knowledge and observing results of experiments using your hypothesis.

A definition of Inductive proof and Deductive proof in mathemetics:

Inductive proof- To proof a mathemetical statement by using an assumption and verifying that the assumption is correct.

Deductive proof- To proof a mathemetical statement by using a series of algebraic rules in order to change the form of one side of the equals sign into the other side and thereby proof an identity.

Now these are not quoted, just digested from what I understood in both subjects. Aren't they contradictions? If you use an assumption then it is a deductive approach whilst if you just follow a path using algebra and observation of how the form changes until you arrive to your answer then it is inductive. Am I wrong?

Or is it inductive because you have no pre defined knowledge and are observing the results given by an assumption whilst the other is deductive because you are using predefined mathematical proofs?
 
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  • #2
Deduction (math and science) is the process of drawing conclusions from hypothesis through a series of logicallly valid steps.

Induction (science) is drawing conclusion based on a series of observations about many different cases of the situation under consideration.

Induction (math) is a completely different thing from that of science. It is a special tool which works like this. Prove something true for n=1. Assume it is true for any n and prove it true for n+1. (Prove meaning using deductive logic). Effectively what happened is that truth for n=1 implies truth for n=2 implies truth for n=3, etc.
 
  • #3
Oh so induction in Maths has nothing to do with science?

Induction always confused me because just because you proof that something is true for n+1, what make you so sure that it is true for n as well?

Also the way you prove that n+1 is true is by making it equal to the assumption + 1. So what?
 
  • #4
mtantu said:
Inductive proof- To proof a mathemetical statement by using an assumption and verifying that the assumption is correct.

Deductive proof- To proof a mathemetical statement by using a series of algebraic rules in order to change the form of one side of the equals sign into the other side and thereby proof an identity.
Well, these are simply nonsense. Inductive Reasoning (not "inductive proof") involves reasoning from a number of special examples to a general principal- but it is never "proof".

"Deductive reasoning" involves reasoning from a general principal to a special case- but it does not necessarily have to be "algebraic" nor is it only used to prove "identities"!
 
  • #5
mtanti said:
Oh so induction in Maths has nothing to do with science?

Induction always confused me because just because you proof that something is true for n+1, what make you so sure that it is true for n as well?
You prove (not "proof"- that's the noun) first that the statement is true for n= 1 so you know it is true for some n- and that's what you are using: if it is true for some n, then it is true for n+1.

Also the way you prove that n+1 is true is by making it equal to the assumption + 1. So what?
I have no idea what you mean here. Making what equal to "the asssumption +1"??
 
  • #6
In more casual terms, induction is taking a finite number of observations and "inducing" a larger conclusion. Proof by induction is called "proof by induction" because it requires only two observations: that it is true for at least one number, and then if it's true for that one number, then it must be true for the next, and the next, and so on. As a more shaky example of inductive reasoning: "it looks like a duck, it quacks like a duck, it's a duck."

Deduction, on the other hand, is taking a general law and "deducing" lesser conclusions. For example, if we know that for any two numbers a and b, ab = ba, then 3*5 = 5*3. Deductive reasoning is always reliable; inductive reasoning isn't usually reliable, for obvious reasons.

Proof by induction requires two steps: first you must make sure that the statement to be proved is true for at least one number -- for our purposes, call it n0 (it doesn't have to be 0). Then, you show that if the statement holds for any number, then it's true for the next number. Only by combining these two steps is an inductive proof complete. Therefore, it must be true for every integer greater than n0.

Proof by induction, unlike most inductive reasoning, is always reliable. It's kind of a special case of inductive reasoning.
 
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  • #7
mtanti said:
Induction always confused me because just because you proof that something is true for n+1, what make you so sure that it is true for n as well?

Also the way you prove that n+1 is true is by making it equal to the assumption + 1. So what?

I think you misunderstand how proof by induction works. I try to explain:

Part A:
The crucial part is to show the step "S(n) -> S(n+1)". In this step we assume that a certain statement S is true for some number n (We write: S(n) is true). It is really important to understand that we only ASSUME that. We do NOT prove it.
Then we use this assumption and show that the statement S is also true for n+1.
In short: IF S(n) true, then we must show that S(n+1) true.

We did NOT prove that statement S is true for some number n, we just ASSUMED it and used it to show some other statement. It's just like saying: IF x>0, then 3*x>0.

Part B:
And now comes the second part that you have to understand. Suppose you have shown that from S(n) it follows S(n+1). Just do the following: show that your statement is true for n=1, that is S(1). After you've done that, what can you conclude?

Well, IF S(1) true, then S(2) is true. (This is because you have shown in Part A: IF S(n) true, then S(n+1) true).
Now, IF S(2) is true, S(3) is true. And so on...
You can climb up to any arbitrary n and show that statement S is true.
This is some sort of domino effect. If the first domino falls, the second does, the third does and so on.

Note: Usually, you first show Part B, that is you first plug in some value for n, for example n=1 and show that S(1) is true. If S(1) is not true, you know that the "first domino" does not fall.
 
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  • #8
good explanation, however how do you prove that if S(n) is true then S(n+1) is also true? Thats the confusing part...

Like I said which perhaps was not clear... "Also the way you prove that n+1 is true is by making it equal to the assumption + 1. So what?" means that you prove S(n+1) is true by transforming it into S(n)+1. OK it won't be an actual +1 but S(n+1) must be expressed as a function of S(n), your assumption (perhaps I'm forgetting what comes next...). So what good will that do in order to prove that if S(n) is true then S(n+1) is also true?
 
  • #9
mtanti said:
however how do you prove that if S(n) is true then S(n+1) is also true?
Start with the hypothesis that S(n) is true.
Derive some stuff.
Arrive at the conclusion that S(n+1) is true.


Here's a simple example.

Suppose I tell you that f(1) = 0, and that f(n + 1) always equals f(n). How would you prove that, for any integer n > 0, f(n) = 0?
 
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  • #10
Hurkyl said:
Suppose I tell you that f(1) = 0, and that f(n + 1) always equals f(n). How would you prove that, for any integer n > 0, f(n) = 0?

Well you see what is being done to the number in the function I guess, such as looking for a simplification which is f(n)=n-1

So how do you arrive at your conclusion that s(n+1) is true? You are basing your proof on an assumption after all...
 
  • #11
Oddly enough mathematical induction is actually a form of deduction. Part of the rigor of mathematics is conclusions follow from previously proven statements or axioms by necessity. This is not the case with scientific induction. I like the set approach to mathematical induction more because it seems to make more apparent the deductive nature of mathematical induction.

Let us say we have some statement, namely P(n). What we want to know is which n’s will make P(n) true. So let us consider the set of n that make P(n) true. We will call this set “S”. First we show that 1 is an element of S. Second we show if n is an element of S, then n+1 must also be an element of S. But, then the natural numbers must be a subset of S, according to the definition of the natural numbers. Thus the natural numbers satisfy P(n).

This might sound a bit sketchy so let’s follow the induction for awhile to see how it pans out.

We start by considering the set S of n’s that make P(n) true. We prove that 1 is an element of S. Then we prove if n is in S n+1 is in S*. So we know 1 is in S because that is the first thing we prove. But by * if 1 is in S, 2 must be in S also. But again by * if 2 is in S then 3 is there also. Yet again by * if 3 is in S then 4 is too. This process can be carried out indefinitely. Given any natural number, I can repeat this process some number of times and show that that number is in the set S.
 
  • #12
mtanti said:
Well you see what is being done to the number in the function I guess, such as looking for a simplification which is f(n)=n-1

So how do you arrive at your conclusion that s(n+1) is true? You are basing your proof on an assumption after all...
Why don’t we try mathematical induction on another statement and see if it helps.

Let P(n) be the statement: The sum of all the natural numbers from 1 to n is equal to n(n+1)/2

Let’s start by considering all the n’s that will make this statement true. We will call the set of these n’s S.

Claim: 1 is in S.
Proof: (1)(1+1)/2 = (1)(2)/2 = 2/2 = 1

Claim 2: if any natural number n is in S, n+1 is also in S
Proof: Let us assume that n is in S. That is the sum of all the natural numbers from 1 to n is equal to n(n+1)/2. Thus…

1+2+3+4+5+…+n+(n+1) =
n(n+1)/2 + (n+1) =
n(n+1)/2 + 2(n+1)/2 =
[n(n+1) + 2(n+1)]/2 =
[n^2 + n + 2n+2)]/2 =
[n^2 + 3n + 2)]/2 =
[(n+1)(n+2)]/2
Thus n+1 satisfies P(n)

Notice going from the first to the second line is justified by our assumption. We ASSUME that P(n) is true for n. That is the sum of all the natural numbers from 1 to n is equal to n(n+1)/2. And we use that to then prove n+1 is true, that is: The sum of all the natural numbers from 1 to n+1 is equal to (n+1)(n+1+1)/2

But by claim 1 and claim 2 we have now proven that the natural numbers are a subset of S by the definition of the natural numbers.
 
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  • #13
OK OK, I've got the actual question now... Sorry but it's not easy to come up with the right question in maths...

---------------------------
Why is it that just because you manage to simplify P(n)+([n+1]th term) into P(n+1) mean that P(n) is true?
---------------------------

Is there a mathemathical reason for this or am I doubting reality and any kind of proof?
 
  • #14
mtanti said:
Why is it that just because you manage to simplify P(n)+([n+1]th term) into P(n+1) mean that P(n) is true?

After simplifying "P(n)+([n+1]th term) into P(n+1)" we have ONLY shown:
IF P(n) is true for a SINGLE n, then P(n+1) is also true.

This step does not prove that P(n) is true for ALL n!

Is that maybe the part that is confusing?
 
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  • #15
No the part that's confusing is why is it that after simplifying "P(n)+([n+1]th term) into P(n+1)" you have proved that P(n+1) is true if P(n) is true. What makes you so sure?
 
  • #16
Hurkyl said:
Suppose I tell you that f(1) = 0, and that f(n + 1) always equals f(n). How would you prove that, for any integer n > 0, f(n) = 0?
mtanti said:
Well you see what is being done to the number in the function I guess, such as looking for a simplification which is f(n)=n-1

So how do you arrive at your conclusion that s(n+1) is true? You are basing your proof on an assumption after all...
If you can't see how to prove it with mathematical induction, can you come up with some other way to prove it? (even if it's just an intuitive proof?)


also, f(n) = n-1 is wrong. (Since f(2) = 0)

Most proofs are based on an assumption -- you assume that the hypothesis of your postulate is true, and from that you derive the conclusion of your postulate.


or, more formally, when P is false, "P => Q" is true.
 
  • #17
mtanti said:
No the part that's confusing is why is it that after simplifying "P(n)+([n+1]th term) into P(n+1)" you have proved that P(n+1) is true if P(n) is true. What makes you so sure?

I see your worry here. I think.

OK, so let's do this with the help of a similar example. Say your sitting on a chair one day just playing around with numbers when you, amazingly enough, "discovered" this relationship:

1 + 2 + 3 + ... n = n(n+1)/2

You were surprised. You checked if the above relationship is true for the numbers 1,2,3,4, and 5. But you know that you can't possibly check if this is true for ALL numbers! Life is too short for that.

You are wondering what to do, when you suddenly recall this beast called "mathematical induction." You are extremely happy and now you set to prove your relationship.

You start with the base case of n=1. You check.
Left hand side: Duh! Obviosly the sum of all number from 1 to 1 is 1.
Right hand side: 1(1+1)/2 = 1
Marvelous! You proved that your relationship is true for n = 1.

Now for the hard part. You know that you have to prove that P(n) => P(n+1) (=> meaning implies) (meaning if P(n) is true then P(n+1) HAS TO BE TRUE). If you can do this, then since you relationship is true for n=1 it HAS TO BE TRUE for n=2 and since its true for n=2 it HAS TO BE TRUE for n=3 and since its true for n=3 ... and so on.

You are determined to prove that P(n) => P(n+1). You make the statement P(n) the relationship you found earlier i.e.

1 + 2 + 3 + ... n = n(n+1)/2 (Note that this WHOLE statement is your P(n) )

Now, just to be conservative, you write down P(n+1) as well i.e.

1 + 2 + 3 + ... + n + n+1 = (n+1)(n+1+1)/2 = (n+1)(n+2)/2

You know that you have to prove that P(n+1) LOGICALLT FOLLOWS from P(n). You look at P(n+1). The left hand side of P(n+1) looks just like P(n) except that it has an extra n+1 term. What you do? Your aim is to DERIVE P(n+1) FROM P(n). So you do what you think is best. You add the term n+1 to both sides of the satement P(n). You know that there is noting wrong with adding the same quantity to both sides of an equation. After you do that, your P(n) looks like the following

1 + 2 + 3 + ... n + (n+1) = n(n+1)/2 + (n+1)

Now, if only you could prove that
n(n+1)/2 + (n+1) = (n+1)(n+2)/2
then you would have successfully derived P(n+1) from P(n).

You go do the algebra and you successfully manage to transform the left hand side of the above equation (i.e. n(n+1)/2 + (n+1) ) into (n+1)(n+2)/2. You have proved that
n(n+1)/2 + (n+1) = (n+1)(n+2)/2
and thus P(n) => P(n+1)
(actually, P(n) and P(n=1) are "equal" but but I am not sure how the word "equal" applies to statements. "equavalent" is a better word)

... and the rest is history. :biggrin:
 
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  • #18
mtanti said:
No the part that's confusing is why is it that after simplifying "P(n)+([n+1]th term) into P(n+1)" you have proved that P(n+1) is true if P(n) is true. What makes you so sure?
What you have written here is somewhat misleading. Its not that you are simplifing "P(n)+([n+1]th term)" into "P(n+1)." You are simplifying "P(n)" into "P(n+1)." It doesn't really makes sense to ADD "([n+1]th term)" to "P(n)" since you are adding the term n+1 to both sides of the statement P(n) and thus you are not changing P(n) in any way.

Hope this helps. :smile:
 
  • #19
Also, just for your information, mathematical induction can only be used to proved relationships which involve whole numbers. However, even though mathematical induction is somewhat limit, nevertheless, it an extremely useful "trick" in mathematics.
 
  • #20
You guys are losing me. To pick on someone at random...
Hurkyl said:
Start with the hypothesis that S(n) is true.
That seems like saying "10 Oranges is true". It doesn't make any sense. What am I missing?
 
  • #21
WhyIsItSo said:
That seems like saying "10 Oranges is true". It doesn't make any sense. What am I missing?

S(n) represents a statement about whole numbers. For exemple, if we want to prove that

[tex]\sum_{i=1}^n i = \frac{n(n+1)}{2}[/tex],

S(n) represents the above equality. The statement S(9) is true means that the equality

[tex]\sum_{i=1}^9 i= \frac{9(9+1)}{2}[/tex]

is true.

The principle of mathematical induction is based on the following logic:

If we can prove that S(n+1) is true based only on the assumption that S(n) is true, and if we can prove that S(1) is true, then it follows that S is true for any integer.

Why? Because we've shown that if S(1) is true, then, S(1+1)=S(2) is true. And this implies in its turn that S(2+1)=S(3) is true. etc.
 
  • #22
WhyIsItSo said:
You guys are losing me. To pick on someone at random...

That seems like saying "10 Oranges is true". It doesn't make any sense. What am I missing?

What are you talking about??

"The statement S(n) is true" means that the right hand side and the left hand side of S(n) are equal for any arbitrary value of n (in the proper domain that is).

What's wrong with saying, "Start with the hypothesis that S(n) is true"?
 
  • #23
In my opinion, the confusion here is in the notation S(N). S(N) isn't a typical function. S(N) is in fact more like a statement. But an example will illustrate the idea much better than vague statements like that.

In our example, S(N) means "The number N has the property that if you sum up the numbers (i.e. integers) from 1 to N, that sum will be equal to (N)(N+1)/2". In other words, S(N) is either yes or no, or true or false if you prefer. That's why we write it S(N) -- because whether or not S is true or false depends on N.

Proof by induction first shows that S(1) is true. That is to say the following: "The number N has the property that if you sum up the numbers from 1 to 1, that sum will be equal to (1)(1+1)/2". Well, that equality holds, because the sum is 1 and (1)(1+1)/2 = 2/2 = 1, so they're equal. Therefore, S(1) is true.

Now, the deal for proof by induction is that we must show is that if S(N) is true, then S(N+1) must also be true. What that means is that if:

"N has the property that the sum of numbers from 1 to N = (N)(N+1)/2"

then it must also be true that

"N+1 has the property that the sum of numbers from 1 to N+1 = (N+1)(N+2)/2".

A little algebra shows this to be true, algebra which I won't repeat here because it's been done above.

Then you get the domino effect also described above, and QED!
 
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  • #24
Hello again, So if you can attain S(N+1) by expanding S(N) than the assumption was correct for some set of values? But why is that?

The real deal with the problem is that you proof is still based on an assumption... If you get S(N+1) from S(N) doesn't really mean anything because S(N) is just an assumption. OK so here is the master question:

Is it possible to attain S(N+1) from S(N) by expansion if S(N) is wrong?
Is it possible to not attain S(N+1) from S(N) by expansion if S(N) is correct?
 
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  • #25
mtanti said:
Hello again, So if you can attain S(N+1) by expanding S(N) than the assumption was correct for some set of values? But why is that?
I am not sure if you are understanding the logic behind mathematical induction. Your question doesn't make any sense.

mtanti said:
The real deal with the problem is that you proof is still based on an assumption... If you get S(N+1) from S(N) doesn't really mean anything because S(N) is just an assumption.
I think you should read this discussion one more tme, because this question has been answered many times (in many different ways).
Let me try to explain one last time,
After you derive S(n+1) from S(n), you come up with the following conclusion: "IF S(n) is true, then S(n+1) is true" (let me call this (*) )
Yes, S(n) is still an assumption (that's why we have the qulifier IF).
BUT, for the base case we proved that S(1) is true. From (*) we can conclude that since S(1) is true therefore S(2) is true, then since S(2) is true, it follows from (*) that S(3) is also true, and so on till infinity.

mtanti said:
OK so here is the master question:

Is it possible to attain S(N+1) from S(N) by expansion if S(N) is wrong?
Is it possible to not attain S(N+1) from S(N) by expansion if S(N) is correct?
I will answer this later. I got to run.
 
  • #26
mtanti said:
The real deal with the problem is that you proof is still based on an assumption... If you get S(N+1) from S(N) doesn't really mean anything because S(N) is just an assumption.
Yes, you've shown that assuming S(N) to be true, S(N+1) is true. That means that if ever we find a number m for which S(m) is true, it will be true for S(m+1) also. And for S([m+1]+1)=S(m+2) also by the same argument, etc.
 
  • #27
mtanti said:
Is it possible to attain S(N+1) from S(N) by expansion if S(N) is wrong?
Is it possible to not attain S(N+1) from S(N) by expansion if S(N) is correct?
I am not sure exactly what you are asking. But anyways, I think that your worry here is that induction might be used to prove things that are false! But that's not the case, trust me.

Basically, you need two things for induction to work. First is the base case and the second is the inductive case. For the base case you have to prove that S(a) is true where "a" is any number in the proper domain (which is generally the domain of natural numbers). For the inductive case, you have to prove that S(n) => S(n+1).

So you need two things:
1) S(a) is true (for any number a)
2) S(n) => S(n+1) is true.

Technically, it is possible to fulfill either step 1) or step 2) or both or none. But your PROOF will only be valid if you fulfill both steps. So even if you can prove step 2), but can never find a value "a" such that S(a) is true, your proof is incomplete.
 
  • #28
I think your difficulty is about logical implication. The statement if X then Y means something very specific to logic and mathematics. If you say If X then Y (denoted X=>Y) what you are claim is that if X is true, then Y must follow no mater what. But note, this itself makes NO CLAIM about the truth value of X and Y. X=>Y is not saying X is true, it also isn’t saying Y is true, what it is saying is there is a relationship between X and Y’s truth values.

So let’s say we prove P=>Q, what does this mean?

First it means if P is true, Q is true also. But note: we still are obligated to prove P is true. What if P is false? Then our “P=>Q” statement tells us nothing about Q. What if we somehow prove that Q is false, then our “P=>Q” tells us that that P is also false. This may be a bit confusing, but the key to understanding implication is that when we claim P=>Q we are affirming a relationship between P and Q, not P or Q themselves.

So when we prove S(n) => S(n+1) in our induction what we are actually proving is there is some relationship between S(n) and S(n+1) namely that if S(n) is true, S(n+1) must also be true. We aren’t claiming anything at all about the truth value of S(n).
 
  • #29
First of all all I am speaking about is the inductive step, not the base case step. Secondly I am not saying that if S(n) => S(n+1) then the assumption is true, I am just talking about the step.

So what are you showing exactly when you prove the inductive step alone? Out of the proof by induction context... What does it mathematically mean to show that you can expand S(N) into S(N+1)?

I know this is getting boring but this thread seems to be getting interest just the same! :)
 
  • #30
Why do you say "expand"?

The essence of induction is contained in two steps:

1) Prove S(1) is true.
2) Prove S(N) true ==> S(N+1) true.

Now if you ask "what does it mean mathematically to show that "S(N) true ==> S(N+1) true", it means just that: that is S is true for one interger, it is true for the next integer as well. This, together, with the fact that S(1) is true means that S(2) is true, which means that S(3) is true, etc.

If you agree to let go of the concept of "expanding S(N) into S(N+1)" and replace it by the concept of "proving S(N+1) is true under the assumption that S(N) is true", is there something you don't understand behind the logic of induction? If you do, I fear there's nothing more we can do for you. If you really want to get this, I'd recommand you take a break trying to understand it. Then come back to the thread in a few weeks, read it all over again with an open mind, and actually practice doing proofs by inductions for yourself. This is where the real insight is gained.
 
  • #31
The whole question circles around how you prove that S(N) true ==> S(N+1) true
 
  • #32
You just use your ingenuity and try to find a way. For instance, for the statement

[tex]\sum_{i=1}^n i = \frac{n(n+1)}{2}[/tex]

The statement S(N) is,

[tex]\sum_{i=1}^N i = \frac{N(N+1)}{2}[/tex]

While S(N+1) is,

[tex]\sum_{i=1}^{N+1} i = \frac{(N+1)(N+2)}{2}[/tex]

Assuming S(N) is true (i.e. assuming that the equality btw the left hand side and the right hand side holds), let us try to prove that S(N+1) is also true.

Let us add the number N+1 on both sides of the equation for S(N). It becomes,

[tex](N+1)+\sum_{i=1}^N i = \frac{N(N+1)}{2}+(N+1) \ \ \ \mbox{(*)}[/tex]

We can "incorporate" the (N+1) in the sumation notation in the following way:

[tex](N+1)+\sum_{i=1}^N i = \sum_{i=1}^{N+1} i[/tex]

While we can manipulate the right hand side this way:

[tex]\frac{N(N+1)}{2}+(N+1) = (N+1)\frac{N}{2}+(N+1) = (N+1)\left[ \frac{N}{2}+1 \right] = (N+1)\left[ \frac{N}{2}+\frac{2}{2} \right]=(N+1)\left[ \frac{N+2}{2}\right] = \frac{(N+1)(N+2)}{2}[/tex]

Such that equation (*) can we rewriten as,

[tex]\sum_{i=1}^{N+1} i = \frac{(N+1)(N+2)}{2}[/tex]

which is precisely the statement S(N+1).
 
  • #33
And this is a deductive proof right? So you prove that, assuming truth for S(N), S(N+1) is also true by deductive method? Because deductive proof works by showing that the left hand side is equal to the right hand side, and that is what you are doing when you show that

S(N+1) = (N+1)((N+1)+1)/2

Because N+1 fitted into the equation, it means that you can find S(N+1) too, assuming that S(N) is true. And you show that S(N) is true by finding a value which shows it is true.

Cool...
 
  • #34
I don't follow most of your reasoning but it sounds like you're getting a hang of it. Your turn now. Prove the following by induction:

[tex]\sum_{i=1}^{n}(2i-1)=n^2 \ \ \ \ \forall n\in\mathbb{N}[/tex]

(The sum of the n first odd positive integers is n²)
 
Last edited:
  • #35
Hey man, I know how to prove by induction, I just don't know what I'm doing. Is it deductive or not?

RTP S(2n-1) = n^2

STEP 1: Prove that statement is true for n=1
2(1)-1 = 1^2
1 = 1
QED

STEP 2: Prove that if statement is true for n, it is also true for n+1
RTP S(2n+1) = (n+1)^2 [Is this what this step is all about?]

S(2n+1) = n^2 + n+1
RHS cannot be transformed into (n+1)^2, therefore S(n+1) will not work, therefore S(n) is not true.

QEDLESS
 
<h2>1. What is the difference between induction and deduction?</h2><p>Induction is a reasoning process that involves making generalizations based on specific observations or examples. Deduction is a reasoning process that involves drawing specific conclusions from general principles or premises.</p><h2>2. Which approach is considered to be more reliable?</h2><p>Both induction and deduction have their strengths and weaknesses, and neither can be considered to be universally more reliable. It ultimately depends on the specific situation and the type of question being asked.</p><h2>3. How do scientists use induction and deduction in their research?</h2><p>Scientists often use a combination of both induction and deduction in their research. They may use induction to generate hypotheses based on observations, and then use deduction to test those hypotheses and draw conclusions.</p><h2>4. Can induction and deduction be used together?</h2><p>Yes, induction and deduction are not mutually exclusive and can be used together in the scientific method. Induction can be used to generate hypotheses, and then deduction can be used to test those hypotheses and draw conclusions.</p><h2>5. Which approach is more commonly used in scientific research?</h2><p>Both induction and deduction are commonly used in scientific research, but the specific approach used may vary depending on the field of study and the research question being addressed.</p>

1. What is the difference between induction and deduction?

Induction is a reasoning process that involves making generalizations based on specific observations or examples. Deduction is a reasoning process that involves drawing specific conclusions from general principles or premises.

2. Which approach is considered to be more reliable?

Both induction and deduction have their strengths and weaknesses, and neither can be considered to be universally more reliable. It ultimately depends on the specific situation and the type of question being asked.

3. How do scientists use induction and deduction in their research?

Scientists often use a combination of both induction and deduction in their research. They may use induction to generate hypotheses based on observations, and then use deduction to test those hypotheses and draw conclusions.

4. Can induction and deduction be used together?

Yes, induction and deduction are not mutually exclusive and can be used together in the scientific method. Induction can be used to generate hypotheses, and then deduction can be used to test those hypotheses and draw conclusions.

5. Which approach is more commonly used in scientific research?

Both induction and deduction are commonly used in scientific research, but the specific approach used may vary depending on the field of study and the research question being addressed.

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